$\frac{\left(2x-1\right)\left(x+2\right)}{\left(x+3\right)\left(x-1\right)}=0$
$6x^2+47x=8$
$\lim\:_{x\to\:\infty\:}\left(\frac{3x^2+2x}{5x^3-3x^2}\right)$
$\left(w-7\right)\left(w-5\right)$
$\frac{x^5+x^3-x}{x+1}$
$\left(x-2m\right)^3$
$2-x=4x-8$
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